Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int DN=3005,M=1e9+7;
double n,a[DN],dp[DN],sum,p1,p2;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    	cin>>a[i];
    dp[0]=1;
    for(int h=1;h<=n;h++)
    {
    	p1=a[h];
    	p2=1-a[h];
    	for(int i=n;i>=0;i--)
    	{
    		dp[i+1]+=dp[i]*p1;
    		dp[i]*=p2;
    	}
    }
    for(int i=n;i>n/2;i--)
    	sum+=dp[i];
    cout<<fixed<<setprecision(12)<<sum;
}